Question: $\dfrac{ -10p - 8q }{ 3 } = \dfrac{ 7p + r }{ -10 }$ Solve for $p$.
Multiply both sides by the left denominator. $\dfrac{ -10p - 8q }{ {3} } = \dfrac{ 7p + r }{ -10 }$ ${3} \cdot \dfrac{ -10p - 8q }{ {3} } = {3} \cdot \dfrac{ 7p + r }{ -10 }$ $-10p - 8q = {3} \cdot \dfrac { 7p + r }{ -10 }$ Multiply both sides by the right denominator. $-10p - 8q = 3 \cdot \dfrac{ 7p + r }{ -{10} }$ $-{10} \cdot \left( -10p - 8q \right) = -{10} \cdot 3 \cdot \dfrac{ 7p + r }{ -{10} }$ $-{10} \cdot \left( -10p - 8q \right) = 3 \cdot \left( 7p + r \right)$ Distribute both sides $-{10} \cdot \left( -10p - 8q \right) = {3} \cdot \left( 7p + r \right)$ ${100}p + {80}q = {21}p + {3}r$ Combine $p$ terms on the left. ${100p} + 80q = {21p} + 3r$ ${79p} + 80q = 3r$ Move the $q$ term to the right. $79p + {80q} = 3r$ $79p = 3r - {80q}$ Isolate $p$ by dividing both sides by its coefficient. ${79}p = 3r - 80q$ $p = \dfrac{ 3r - 80q }{ {79} }$